I am stuck on a textbook problem on vectors (a new topic for me). The problem given in my book says:

A man can swim with a speed of 4 km/h in still water. How long does he take to cross a river 1 km wide if the river flows steadily at 3 km/h and he takes his strokes normal to the river current?

The author used the speed of man (4 km/h) to calculate the time taken, but I can't understand that why are we not using the resultant of the speeds of man and river to calculate the time instead? Or, is it that the speed of man is unaffected by the speed of river?

  • $\begingroup$ Can you post the textbook solution? $\endgroup$ – DanDan0101 Sep 27 '19 at 15:05
  • $\begingroup$ Sure, just a minute $\endgroup$ – UserKunal123 Sep 27 '19 at 15:06
  • $\begingroup$ Sorry for taking this long, but I am unable to upload a screenshot, the app says that something went wrong. :( $\endgroup$ – UserKunal123 Sep 27 '19 at 15:19
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    $\begingroup$ I've removed some inappropriate comments, and replies to them. Be nice, people. $\endgroup$ – rob Sep 27 '19 at 18:37

He is going normal (perpendicular) to the river current so the time it takes him to cross it doesn't depend on the river current, he will just be further down the river.

  • $\begingroup$ I get it now. What I was thinking is that wouldn't the river current reduce the speed of the man a bit, and if so, wouldn't the reduced speed be his actual speed? $\endgroup$ – UserKunal123 Sep 27 '19 at 15:26
  • $\begingroup$ The river current changes his speed yes (5 km/h), but not the time it takes him to cross the river. I think this is a bit of a trick question, since it tries to make you think the river current speed matters. $\endgroup$ – user234190 Sep 27 '19 at 15:29
  • $\begingroup$ @UserKunal123 "wouldn't the river current reduce the speed of the man a bit" You are right that the river current will change his speed (it will not reduce it, though, but increase it). But it will only increase his parallel speed (parallel to the current). Not his perpendicular, or normal, speed. And only his normal speed influences any motion in the normal direction. Any speed he might gain along other dimensions do not influence his motion along his headed dimension. That's the point of the question. $\endgroup$ – Steeven Sep 27 '19 at 17:56
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    $\begingroup$ I've removed an inappropriate comment, and a reply to it. $\endgroup$ – rob Sep 27 '19 at 18:39

why are we not using the resultant of the speeds of man and river to calculate the time instead? Or, is it that the speed of man is unaffected by the speed of river?

His (resulting) speed is changed by the river (it will not reduce it, though, but increase it). But his resulting speed is not what you need to solve this. And that is why they don't include it.

Let's imagine a y-axis along the river and an x-axis from shore-to-shore, normal to the river.

  • In still water, he can swim straight across from shore to shore. His velocity is: $$\vec v=(v_x,0)$$ His speed will be $v=\sqrt{v_x^2+0^2}=v_x$. In other words, his speed is the same as the normal component-speed.

We can set up a motion equation in the y- and in the x-direction to calculate what we need. If we set one up in the y-direction, for example this one: $$\require{cancel} s_y=s_{y,0}+\cancel{v_{y,0}}t+\frac12 \cancel a t^2\quad\Leftrightarrow \quad s_y=s_{y,0}$$ then we see that he doesn't move in that direction at all, because his speed (and also acceleration) is zero. You can set something similar up in the x-direction, and there you will get an expression where you can find the time $t$ it takes him to cross.

  • In flowing water, he still swims across, but this time he will be moved along with the river in the y-direction simultaneously. He will acquire a y-speed as well: $$\vec v=(v_x,v_y)$$ His speed will now be $v=\sqrt{v_x^2+v_y^2}$. In other words, his speed $v$ is this time larger than the normal component-speed $v_x$. But, this is actually irrelevant. We are not going to use $v$ anywhere in any equations, we are only going to use $v_x$. Because, the motion equations are directional.

We can again set up a motion equation in the y- and in the x-direction. In the y-direction, we this time get: $$s_y=s_{y,0}+v_{y,0}t+\frac12 \cancel a t^2\quad\Leftrightarrow\quad s_y=s_{y,0}+v_{y,0}t$$ which shows us that his position $s_y$ grows and grows each second. Indeed, he does flow along with the river (assuming he has the same speed as the river current speed). This doesn't help him across the river, though. If he didn't swim in the x-direction, then he would never reach the other shore. This y-speed does not influence his x-motion at all. In fact, if you set up an equation in the x-direction again, then you will see that you just have the same speed as before. You will get the same result as in still water.

He may be pulled along with the river, so he may reach the shore further down the river. But he nevertheless reaches the shore in the same time, because the river flow doesn't affect his normal motion at all.


This problem is not stated well, and I think they got the numbers wrong. They meant to say he could swim at 5 km/h, so that when he swims straight across the 3 km/h river [they left that out], he's going 4 km/h miles-made-good. So it takes 15 minutes.

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    $\begingroup$ If you take a physics test and write the answers to the questions you think should be there, rather than the ones that are, good luck with that. $\endgroup$ – user234190 Sep 27 '19 at 15:38
  • $\begingroup$ But, if the river current does reduce his speed, then why are we not using that reduced or resultant speed instead? It's actually still not clear to me. $\endgroup$ – UserKunal123 Sep 27 '19 at 15:47
  • $\begingroup$ I didn't solve the homework problem as stated. I solved what they meant to say. They knew $4^2+3^2=5^2$, but forget the solution to their problem is $4^2-3^2$. Basically, the text book should be put in the shredder, it's junk. Not sure why I got a down vote, I didn't (and wouldn't) write a junk text book. $\endgroup$ – JEB Sep 27 '19 at 16:12
  • $\begingroup$ @JEB Why do you think this question is wrong? I don't see why you would say that. It is definitely solvable with the given info and description. It is a simple $\Delta s=vt$ equation in the normal direction. It just contains a slight "trick" to confuse the student by giving the unnecessary current speed. $\endgroup$ – Steeven Sep 27 '19 at 17:38

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